I am thinking about finding the Zariski closure of this subset of $\Bbb C^2$. If I got it correct, I need to find a minimal algebraic variety that contains $\{(a,0) \;|\; a ∈ \Bbb Z\}$ as a subset. I know that I cannot take infinite union of varieties so I cannot take the union of each single point. So is that the fact that closure is just the real line with the relation $y$? (If so, how may I prove it?) How can I think about it?
Any hint or links are also appreciated. Thanks in advance.
The Zariski closure is just the $x$-axis. All you have to do is prove that if a polynomial $f(x,y)$ has $f(n,0)=0$ for all $n\in\Bbb Z$ then $f(x,0)=0$ for all $x\in\Bbb C$.
Write $f(x,y)=g(x)+yh(x,y)$. Then $g(x)=0$ for all $n\in\Bbb Z$, so it's a polynomial with infinitely many zeroes...